螺旋矩阵.

题目链接

思路

• 分层的去将数组中的元素返回
• 定义上下左右四个临界节点;
• 每次从上左->上右->右下->左下->上左
• 注意:临界值避免多次使用或者不使用

java

public List<Integer> spiralOrder(int[][] matrix) {
    if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
        return new ArrayList<>();
    }
    int left = 0;
    int right = matrix[0].length - 1;
    int top = 0;
    int bottom = matrix.length - 1;
    List<Integer> res = new ArrayList<>((right + 1) * (bottom + 1));
    while (left <= right && top <= bottom) {
        for (int i = left; i <= right; i++) {
            res.add(matrix[top][i]);
        }
        for (int i = top + 1; i <= bottom; i++) {
            res.add(matrix[i][right]);
        }
        if (left < right && top < bottom) {
            for (int i = right - 1; i >= left; i--) {
                res.add(matrix[bottom][i]);
            }
            for (int i = bottom - 1; i > top; i--) {
                res.add(matrix[i][left]);
            }
        }
        left++;
        right--;
        top++;
        bottom--;
    }
    return res;
}

go

func spiralOrder(matrix [][]int) []int {
	var res []int
	if len(matrix) == 0 || len(matrix[0]) == 0 {
		return res
	}
	left, right, top, button := 0, len(matrix[0])-1, 0, len(matrix)-1
	for left <= right && top <= button {
		for i := left; i <= right; i++ {
			res = append(res, matrix[top][i])
		}
		for i := top + 1; i <= button; i++ {
			res = append(res, matrix[i][right])
		}
		if left < right && top < button {
			for i := right - 1; i >= left; i-- {
				res = append(res, matrix[button][i])
			}
			for i := button - 1; i > top; i-- {
				res = append(res, matrix[i][left])
			}
		}
		left++
		right--
		top++
		button--
	}
	return res
}